3 Answers
3

Your description is almost correct. See Nitsure's notes: http://arxiv.org/PS_cache/math/pdf/0504/0504590v1.pdf , you can find it under "elementary examples". These notes are also included in the book "Fundamental Algebraic Geometry: Grothendieck's FGA Explained" which is really great.

On the Grassmannian X = Gr(k,n) (I am using this notation to mean k-dimensional subspaces of an n-dimensional vector space), we have the trivial bundle $X \times K^n$ (here K is our field of definition), and the tautological subbundle R (naively, this is the subset {$(x,v) \in X \times K^n \mid v \in x$} which is a locally free sheaf of rank k, and its quotient Q is also a locally free sheaf of rank n-k. So we write

$0 \to R \to X \times K^n \to Q \to 0$,

which is the tautological sequence. Given a map to $f \colon Y \to X$, where Y is a k-scheme, we can pull back this sequence to get

$0 \to f^*R \to Y \times K^n \to f^*Q \to 0$.

Conversely, given such a sequence, there is a unique map to X which gives this pullback. Naively, over a point y, the fiber of $f^*R$ is a subspace of $K^n$, so we send it to that closed point. So X represents the functor which sends Y to the set of short exact sequences as above. The quotient being locally free implies that the subsheaf is also locally free, so it really represents the functor which sends Y to the set of quotients $\mathcal{O}_Y^n \to F \to 0$ where the rank of F is n-k.

For general flag varieties, we have similar tautological sequences (but there are more subbundles to consider). This kind of analysis also makes sense for symplectic and orthogonal Grassmannians and flag varieties.

We can also do something similar if we're talking about the Grassmannian of a vector bundle instead of a vector space. Then we just work in the category of S-schemes instead of k-schemes where S is whatever the base space is.

When Y is not noetherian, don't we want to replace the condition that the quotient F of O^n_Y is locally free with the condition that it is flat over Y? (I'm thinking here about the formulation of the Quot scheme, which really deserves to be mentioned explicitly somewhere in an answer to this question).
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Peter McNamaraNov 25 '09 at 5:21

Peter: I don't think it matters if Y is Noetherian or not. I picked this stuff up from Aaron Bertram's notes: math.utah.edu/~bertram/courses/hilbert and looking back on it, I guess I was supposed to put an equivalence relation on those short exact sequences.
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Steven SamNov 25 '09 at 6:37

OK so according to those notes you linked to, for any quotient of O^n_Y, being flat with Hilbert polynomial k is equivalent to being locally free of rank k. (Now I just have to work out why this is true)
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Peter McNamaraNov 26 '09 at 15:28

It's better to think of projective $n$-space as representing the functor which sends $X$ to the set of rank $1$ sub-vector bundles of the trivial rank $n+1$ vector bundle over $X$ [such that the quotient is a rank $n$ vector bundle], mod isomorphism.

The Grassmannian $G(r,n)$ represents the functor which sends $X$ to the set of rank $r$ sub-vector bundles of the trivial rank $n$ vector bundle over $X$ [such that the quotient is a rank $n-r$ vector bundle], mod isomorphism.

The flag variety $F(r_1, \dots, r_k, n)$ represents the functor which sends $X$ to the set of tuples $(\mathcal{F}_1, \dots, \mathcal{F}_k)$ such that $\mathcal{F}_1 \subset \cdots \subset \mathcal{F}_k \subset \mathcal{O}_X^{n}$ and such that each $\mathcal{F}_i$ is a vector bundle of rank $r_i$ [and such that each of the corresponding quotients are vector bundles of the appropriate rank], mod isomorphism.

You can generalize further. Fix a scheme $S$ and a vector bundle $\mathcal{E}$ over $S$. Then in the above, with each $X$ being an $S$-scheme, replace the trivial bundles with the pullback of $\mathcal{E}$.

Kevin: if we're working in the algebraic category, this isn't right. You need that the quotient of the subbundle is torsionfree. So it is better to say that the functor represents quotient bundles (because the kernel is automatically locally free).
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Steven SamNov 24 '09 at 21:44

This is not true. There may exist a an injection of locally free sheaves such that the quotient is not torsion free. Still a sub-vector-bundle is contained in the vector bundle fiberwise, so the quotient is always a vector bundle. Said otherwise let E and F be vector bundles. One can have an injective morphism of sheaves between (the sheaf sections of) E and F without E being a sub-vector-bundle of F. This is because a morphism of sheaves is injective iff it is injective on stalks, not on fibers.
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Andrea FerrettiNov 25 '09 at 10:18

Andrea: So in summary, are you saying that what I wrote is correct, and that I can remove the statements that I had added in the square brackets? And you are saying that if my statements had involved locally free sheaves rather than vector bundles, then I would have to keep the statements in the square brackets?
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Kevin H. LinNov 25 '09 at 22:43

Andrea: If I am interpreting your comment correctly, are you thinking of a vector bundle as the fibers of a locally free sheaf? If so, then what you mentioned is fine. In my comment I was implicitly thinking of vector bundles and locally free sheaves as the same thing (since the question was tagged with "algebraic geometry" I thought this seemed reasonable).
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Steven SamNov 26 '09 at 16:37